Do once-punctured torus bundle groups have non-empty ordinary sets? Let $M$ be a once-punctured torus bundle which is a hyperbolic 3-manifold. Associated to such a hyperbolic 3-manifold is a discrete subgroup $\Gamma \subset PSL(2, \mathbb{C})$ (a Kleinian group) isomorphic to $\pi_1(M)$.
$\Gamma$ acts discontinuously on $H^3$, but not on the conformal boundary $\hat{\mathbb{C}} = P^1 \mathbb{C}$ of $H^3$. Rather, $P^1 \mathbb{C}$ decomposes into the closure of an (any) orbit $\Gamma p$, called the limit set $\Lambda(\Gamma)$ of $\Gamma$, and its complement, the ordinary set $\Omega(\Gamma) = P^1 \mathbb{C} \setminus \Lambda(\Gamma)$ of $\Gamma$ on which $\Gamma$ acts discontinuously. See [1], ch. 5.3, for details.
Is it possible for a once-punctured torus bundle that $\Omega(\Gamma) = \emptyset$? I assume this is not the case but do not know where to look for a proof or how to proceed. In [1], Thms. 5.3.14 and 5.3.15, there are criteria for this, but I would not know how to make use of them.
(In the old naming convention, a "Kleinian group" was required to be discrete and satisfy $\Omega(\Gamma) \neq \emptyset$.)
[1] Beardon, Alan F.: The Geometry of Discrete Groups. Graduate Texts in Mathematics vol. 91. Springer (1983). https://doi.org/10.1007/978-1-4612-1146-4.
 A: Not only is it possible, but in fact $\Omega(\Gamma) = \emptyset$ for every $\Gamma$ such that $M = \mathbb H^3/\Gamma$ is a once-punctured torus bundle over a circle. (Beardon's book is mostly about Fuchsian groups, i.e. discrete subgroups of $PSL(2,\mathbb R)$, and it does not contain what you need to know to understand this problem.)
This is a consequence of two theorems.
The first is that if the hyperbolic 3-manifold $M = \mathbb H^3 / \Gamma$ has finite volume then $\Omega(\Gamma) = \emptyset$; you can find this, for example, in Theorem 12.1.15 of Ratcliffe's book.
The second is the "thick-thin" decomposition theorem; see for example section 12.5 of Ratcliffe's book, or section 4.13 of Kapovich's Book. The "thin part" of your manifold $M$ is a $\mathbb Z^2$ cusp, namely a circle bundle over an annulus that surrounds the puncture (that has been removed from the torus), and every $\mathbb Z^2$ cusp has finite volume. The complement of that cusp is a compact subset of $M$ which therefore also has finite volume. Since $M$ is the union of two subsets of finite volume, it has finite volume.
